Question: Divide the following complex numbers. $ \dfrac{-8-4i}{-2+4i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2-4i}$ $ \dfrac{-8-4i}{-2+4i} = \dfrac{-8-4i}{-2+4i} \cdot \dfrac{{-2-4i}}{{-2-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-4i) \cdot (-2-4i)} {(-2+4i) \cdot (-2-4i)} = \dfrac{(-8-4i) \cdot (-2-4i)} {(-2)^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-4i) \cdot (-2-4i)} {(-2)^2 - (4i)^2} = $ $ \dfrac{(-8-4i) \cdot (-2-4i)} {4 + 16} = $ $ \dfrac{(-8-4i) \cdot (-2-4i)} {20} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-4i}) \cdot ({-2-4i})} {20} = $ $ \dfrac{{-8} \cdot {(-2)} + {-4} \cdot {(-2) i} + {-8} \cdot {-4 i} + {-4} \cdot {-4 i^2}} {20} $ Evaluate each product of two numbers. $ \dfrac{16 + 8i + 32i + 16 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{16 + 8i + 32i - 16} {20} = \dfrac{0 + 40i} {20} = 2i $